How do I split a string by whitespace and ignoring leading and trailing whitespace into an array of words using a regular expression?
javascript, regex, removing-whitespace, string, whitespace
Solution
If you are more interested in the bits that are not whitespace, you can match the non-whitespace instead of splitting on whitespace.
" The quick brown fox jumps over the lazy dog. ".match(/\S+/g);
Note that the following returns `null`:
" ".match(/\S+/g)
So the best pattern to learn is:
str.match(/\S+/g) || []
Problem
I typically use the following code in JavaScript to split a string by whitespace. ``` "The quick brown fox jumps over the lazy dog.".split(/\s+/); // ["The", "quick", "brown", "fox", "jumps", "over", "the", "lazy", "dog."] ``` This of course works even when there are multiple whitespace characters between words. ``` "The quick brown fox jumps over the lazy dog.".split(/\s+/); // ["The", "quick", "brown", "fox", "jumps", "over", "the", "lazy", "dog."] ``` The problem is when I have a string that has leading or trailing whitespace in which case the resulting array of strings will include an empty character at the beginning and/or end of the array. ``` " The quick brown fox jumps over the lazy dog. ".split(/\s+/); // ["", "The", "quick", "brown", "fox", "jumps", "over", "the", "lazy", "dog.", ""] ``` It's a trivial task to eliminate such empty characters, but I'd rather take care of this within the regular expression if that's at all possible. Does anybody know what regular expression I could use to accomplish this goal?